Question

A bullet of mass 100 gram is fired from a gun of mass 20 kg with a velocity of 100 metre per second identify the velocity of recoil of the gun

Answer 1

Recoil in of the gun happens due to conservation of linear momentum.

Mass of gun=20kg

Mass of bullet=100g=100/1000=0.1kg

Recoil velocity =v1

Bullet fired with initial velocity, v2=100m/s

Initially the gun is at rest.

Thus initial velocity v=0

Total initial momentum of gun and bullet =(m1+m2) v=0

Total momentum of gun and bullet after firing =m1v1 + m2v2=20v1+0.1×100

20v1 + 10

Law of conservation of linear momentum :

Total momentum after firing = total momentum before firing.

20v1 + 10=0

v1=-10/20=-0.5m/s

Recoil velocity =-0. 5m/s

The negative sign indicates the gun recoil backwards.

Answer 2

Data:
Mass of bullet = 100 g = 100/1000 = 0.1 kg
Mass of a gun = 20 kg
Velocity of bullet = 100 m/sec
Recoil velocity of the gun = ?
Solution:
Momentum of the system before firing = M x 0 + m x 0
                                                              = 0
Momentum of the system after firing    = MV + mv
                                                             = (20) (V) + (0.1) (100)
                                                             = 20V + 10
Since,
Initial Momentum = Final Momentum
So,
20 V + 10 = 0
20 V = - 10
V = - 10/20
V = - 0.5 m/sec
which is the required recoil velocity of the gun.
Hopefully it is helpful.
Thanks.